0.4-kg mass is attached to a spring that can compress as well as stretch. The ma

Question

0.4-kg mass is attached to a spring that can compress as well as stretch. The mass and spring are resting on a horizontal tabletop. The spring constant is 50 N/m. The mass is pulled, stretching the spring 48 cm. The mass is then released, and the spring-mass system begins to oscillate. Predict the speed of the mass as it passes a point that is a distance of 39 cm from its equilibrium point on the other side of the equilibrium position (spring is compressed).

I solved this myself, but got a really big number that doesn't make sense. Please work it out step by step so that I can see what I did wrong and fix it. I greatly appreciate it!!

Explanation / Answer

F = -kx

Hence

F = 50 * 0.48

F = 24 N

Now F = ma

Hence

24 = 0.4* a

a = 60 m/sec^2

s = 0.09 m

hence

v^2 = 0 + 2*60*0.09

v = 3.28 m/s

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